Mixing
On $sup|varphi^{-1}(n)|=+infty$
$begingroup$
I am trying to find an elementary proof of the following fact:
Given some $Ngeq 2$, there are $N$ distinct integers $a_1,ldots,a_N$ such that $varphi(a_1)=ldots=varphi(a_N)$ with $varphi$ being Euler's totient function.
My original analytic proof goes as follows:
If the number of solutions of $varphi(x)=N$ were bounded, the series
$$ sum_{ngeq 2019}frac{1}{varphi(n)log^2varphi(n)}$$
would be convergent by comparison with $sum_{ngeq n_0}frac{1}{n log^2 n}$ and condensation. It is enough to show that the last series is divergent. It is bounded below by a multiple of
$$ sum_{ngeq 2019}frac{sigma(n)}{n^2 log^2(n)}$$
and since
$$ sum_{ngeq 1}frac{sigma(n)}{n^{2+s}}=zeta(s+1)zeta(s+2)$$
for any $s>0$, it is enough to prove that the integral
$$ int_{0}^{+infty}int_{0}^{+infty}zeta(1+t+s)zeta(2+t+s),ds,dt $$
is divergent, or that the integral
$$ int_{0}^{+infty} u,zeta(1+u)zeta(2+u),du $$
is divergent. On the other hand this is trivial since $uzeta(1+u)zeta(2+u)geq u$ for any $u>0$.
Alternative combinatorial proof: we may consider a very large $N$ and the numbers in $[N,2N]$ with at least $logloglog N$ prime factors. They have a positive density in $[N,2N]$, and they are mapped by the totient function into an interval with length $Oleft(frac{N}{log log N}right)$. By the pigeonhole principle, at least $Omega(loglog N)$ elements of $[N,2N]$ share the same $varphi$.
I would be happier in having a combinatorial proof possibly not relying on subtle statements about the average order of $omega(n)$ or Mertens' theorem about $sum_{pleq x}frac{1}{p}$.
number-theory alternative-proof pigeonhole-principle arithmetic-functions dirichlet-series
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
|
show 13 more comments
$begingroup$
I am trying to find an elementary proof of the following fact:
Given some $Ngeq 2$, there are $N$ distinct integers $a_1,ldots,a_N$ such that $varphi(a_1)=ldots=varphi(a_N)$ with $varphi$ being Euler's totient function.
My original analytic proof goes as follows:
If the number of solutions of $varphi(x)=N$ were bounded, the series
$$ sum_{ngeq 2019}frac{1}{varphi(n)log^2varphi(n)}$$
would be convergent by comparison with $sum_{ngeq n_0}frac{1}{n log^2 n}$ and condensation. It is enough to show that the last series is divergent. It is bounded below by a multiple of
$$ sum_{ngeq 2019}frac{sigma(n)}{n^2 log^2(n)}$$
and since
$$ sum_{ngeq 1}frac{sigma(n)}{n^{2+s}}=zeta(s+1)zeta(s+2)$$
for any $s>0$, it is enough to prove that the integral
$$ int_{0}^{+infty}int_{0}^{+infty}zeta(1+t+s)zeta(2+t+s),ds,dt $$
is divergent, or that the integral
$$ int_{0}^{+infty} u,zeta(1+u)zeta(2+u),du $$
is divergent. On the other hand this is trivial since $uzeta(1+u)zeta(2+u)geq u$ for any $u>0$.
Alternative combinatorial proof: we may consider a very large $N$ and the numbers in $[N,2N]$ with at least $logloglog N$ prime factors. They have a positive density in $[N,2N]$, and they are mapped by the totient function into an interval with length $Oleft(frac{N}{log log N}right)$. By the pigeonhole principle, at least $Omega(loglog N)$ elements of $[N,2N]$ share the same $varphi$.
I would be happier in having a combinatorial proof possibly not relying on subtle statements about the average order of $omega(n)$ or Mertens' theorem about $sum_{pleq x}frac{1}{p}$.
number-theory alternative-proof pigeonhole-principle arithmetic-functions dirichlet-series
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
1
$begingroup$
@Adam: what is the open problem you are talking about? I believe my analytic proof clearly shows/proves that $|varphi^{-1}(n)|$ is not bounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:51
1
$begingroup$
Carmicheal's conjecture states a different thing, i.e. that every value in the range of $varphi$ is attained at least twice. I am just stating that there are values attained with arbitrary multiplicity.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:56
1
$begingroup$
@JackD'Aurizio Ford's theorem states that for any $k$, there is $n$ with $|varphi^{-1}(n)|=k$, while your question only asks about $geq k$. I haven't looked up the proof of Ford's theorem, but I expect your problem can be solved in a much simpler way (and might have been known historically earlier)
$endgroup$
– Wojowu
Jan 3 at 23:17
1
$begingroup$
@Wojowu: fine. Here it is another overkill: by Pillai's theorem, the set of values taken by the $varphi$ function has density zero. It follows that $|varphi^{-1}(n)|$ is clearly unbounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 23:43
1
$begingroup$
@JackD'Aurizio typical that no one throws them a log too
$endgroup$
– Adam
Jan 4 at 0:16
|
show 13 more comments
$begingroup$
I am trying to find an elementary proof of the following fact:
Given some $Ngeq 2$, there are $N$ distinct integers $a_1,ldots,a_N$ such that $varphi(a_1)=ldots=varphi(a_N)$ with $varphi$ being Euler's totient function.
My original analytic proof goes as follows:
If the number of solutions of $varphi(x)=N$ were bounded, the series
$$ sum_{ngeq 2019}frac{1}{varphi(n)log^2varphi(n)}$$
would be convergent by comparison with $sum_{ngeq n_0}frac{1}{n log^2 n}$ and condensation. It is enough to show that the last series is divergent. It is bounded below by a multiple of
$$ sum_{ngeq 2019}frac{sigma(n)}{n^2 log^2(n)}$$
and since
$$ sum_{ngeq 1}frac{sigma(n)}{n^{2+s}}=zeta(s+1)zeta(s+2)$$
for any $s>0$, it is enough to prove that the integral
$$ int_{0}^{+infty}int_{0}^{+infty}zeta(1+t+s)zeta(2+t+s),ds,dt $$
is divergent, or that the integral
$$ int_{0}^{+infty} u,zeta(1+u)zeta(2+u),du $$
is divergent. On the other hand this is trivial since $uzeta(1+u)zeta(2+u)geq u$ for any $u>0$.
Alternative combinatorial proof: we may consider a very large $N$ and the numbers in $[N,2N]$ with at least $logloglog N$ prime factors. They have a positive density in $[N,2N]$, and they are mapped by the totient function into an interval with length $Oleft(frac{N}{log log N}right)$. By the pigeonhole principle, at least $Omega(loglog N)$ elements of $[N,2N]$ share the same $varphi$.
I would be happier in having a combinatorial proof possibly not relying on subtle statements about the average order of $omega(n)$ or Mertens' theorem about $sum_{pleq x}frac{1}{p}$.
number-theory alternative-proof pigeonhole-principle arithmetic-functions dirichlet-series
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
I am trying to find an elementary proof of the following fact:
Given some $Ngeq 2$, there are $N$ distinct integers $a_1,ldots,a_N$ such that $varphi(a_1)=ldots=varphi(a_N)$ with $varphi$ being Euler's totient function.
My original analytic proof goes as follows:
If the number of solutions of $varphi(x)=N$ were bounded, the series
$$ sum_{ngeq 2019}frac{1}{varphi(n)log^2varphi(n)}$$
would be convergent by comparison with $sum_{ngeq n_0}frac{1}{n log^2 n}$ and condensation. It is enough to show that the last series is divergent. It is bounded below by a multiple of
$$ sum_{ngeq 2019}frac{sigma(n)}{n^2 log^2(n)}$$
and since
$$ sum_{ngeq 1}frac{sigma(n)}{n^{2+s}}=zeta(s+1)zeta(s+2)$$
for any $s>0$, it is enough to prove that the integral
$$ int_{0}^{+infty}int_{0}^{+infty}zeta(1+t+s)zeta(2+t+s),ds,dt $$
is divergent, or that the integral
$$ int_{0}^{+infty} u,zeta(1+u)zeta(2+u),du $$
is divergent. On the other hand this is trivial since $uzeta(1+u)zeta(2+u)geq u$ for any $u>0$.
Alternative combinatorial proof: we may consider a very large $N$ and the numbers in $[N,2N]$ with at least $logloglog N$ prime factors. They have a positive density in $[N,2N]$, and they are mapped by the totient function into an interval with length $Oleft(frac{N}{log log N}right)$. By the pigeonhole principle, at least $Omega(loglog N)$ elements of $[N,2N]$ share the same $varphi$.
I would be happier in having a combinatorial proof possibly not relying on subtle statements about the average order of $omega(n)$ or Mertens' theorem about $sum_{pleq x}frac{1}{p}$.
number-theory alternative-proof pigeonhole-principle arithmetic-functions dirichlet-series
number-theory alternative-proof pigeonhole-principle arithmetic-functions dirichlet-series
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
asked Jan 3 at 22:38
Jack D'AurizioJack D'Aurizio
288k33280660
288k33280660
1
$begingroup$
@Adam: what is the open problem you are talking about? I believe my analytic proof clearly shows/proves that $|varphi^{-1}(n)|$ is not bounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:51
1
$begingroup$
Carmicheal's conjecture states a different thing, i.e. that every value in the range of $varphi$ is attained at least twice. I am just stating that there are values attained with arbitrary multiplicity.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:56
1
$begingroup$
@JackD'Aurizio Ford's theorem states that for any $k$, there is $n$ with $|varphi^{-1}(n)|=k$, while your question only asks about $geq k$. I haven't looked up the proof of Ford's theorem, but I expect your problem can be solved in a much simpler way (and might have been known historically earlier)
$endgroup$
– Wojowu
Jan 3 at 23:17
1
$begingroup$
@Wojowu: fine. Here it is another overkill: by Pillai's theorem, the set of values taken by the $varphi$ function has density zero. It follows that $|varphi^{-1}(n)|$ is clearly unbounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 23:43
1
$begingroup$
@JackD'Aurizio typical that no one throws them a log too
$endgroup$
– Adam
Jan 4 at 0:16
|
show 13 more comments
1
$begingroup$
@Adam: what is the open problem you are talking about? I believe my analytic proof clearly shows/proves that $|varphi^{-1}(n)|$ is not bounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:51
1
$begingroup$
Carmicheal's conjecture states a different thing, i.e. that every value in the range of $varphi$ is attained at least twice. I am just stating that there are values attained with arbitrary multiplicity.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:56
1
$begingroup$
@JackD'Aurizio Ford's theorem states that for any $k$, there is $n$ with $|varphi^{-1}(n)|=k$, while your question only asks about $geq k$. I haven't looked up the proof of Ford's theorem, but I expect your problem can be solved in a much simpler way (and might have been known historically earlier)
$endgroup$
– Wojowu
Jan 3 at 23:17
1
$begingroup$
@Wojowu: fine. Here it is another overkill: by Pillai's theorem, the set of values taken by the $varphi$ function has density zero. It follows that $|varphi^{-1}(n)|$ is clearly unbounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 23:43
1
$begingroup$
@JackD'Aurizio typical that no one throws them a log too
$endgroup$
– Adam
Jan 4 at 0:16
1
1
$begingroup$
@Adam: what is the open problem you are talking about? I believe my analytic proof clearly shows/proves that $|varphi^{-1}(n)|$ is not bounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:51
$begingroup$
@Adam: what is the open problem you are talking about? I believe my analytic proof clearly shows/proves that $|varphi^{-1}(n)|$ is not bounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:51
1
1
$begingroup$
Carmicheal's conjecture states a different thing, i.e. that every value in the range of $varphi$ is attained at least twice. I am just stating that there are values attained with arbitrary multiplicity.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:56
$begingroup$
Carmicheal's conjecture states a different thing, i.e. that every value in the range of $varphi$ is attained at least twice. I am just stating that there are values attained with arbitrary multiplicity.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:56
1
1
$begingroup$
@JackD'Aurizio Ford's theorem states that for any $k$, there is $n$ with $|varphi^{-1}(n)|=k$, while your question only asks about $geq k$. I haven't looked up the proof of Ford's theorem, but I expect your problem can be solved in a much simpler way (and might have been known historically earlier)
$endgroup$
– Wojowu
Jan 3 at 23:17
$begingroup$
@JackD'Aurizio Ford's theorem states that for any $k$, there is $n$ with $|varphi^{-1}(n)|=k$, while your question only asks about $geq k$. I haven't looked up the proof of Ford's theorem, but I expect your problem can be solved in a much simpler way (and might have been known historically earlier)
$endgroup$
– Wojowu
Jan 3 at 23:17
1
1
$begingroup$
@Wojowu: fine. Here it is another overkill: by Pillai's theorem, the set of values taken by the $varphi$ function has density zero. It follows that $|varphi^{-1}(n)|$ is clearly unbounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 23:43
$begingroup$
@Wojowu: fine. Here it is another overkill: by Pillai's theorem, the set of values taken by the $varphi$ function has density zero. It follows that $|varphi^{-1}(n)|$ is clearly unbounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 23:43
1
1
$begingroup$
@JackD'Aurizio typical that no one throws them a log too
$endgroup$
– Adam
Jan 4 at 0:16
$begingroup$
@JackD'Aurizio typical that no one throws them a log too
$endgroup$
– Adam
Jan 4 at 0:16
|
show 13 more comments
1 Answer
1
active
oldest
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$begingroup$
Here it is an elementary argument inspired by Pillai$^{(*)}$: the point is that $nu_2(varphi(n))approxomega(n).$
Let $O^{omega}$ the set of odd natural numbers with at least $omega$ prime divisors and let $O^{omega}_n=O^{omega}cap[1,n]$.
$O^{omega}$ certainly has density $frac{1}{2}$ for any $omega$ - it is enough to invoke some elementary version of the PNT and basic sieving arguments. The totient function maps $O^{omega}_n$ into a subset of $[1,n]cap 2^{omega}mathbb{N}$, hence by the pigeonhole principle at least $(1-varepsilon_n)2^{omega-1}$ elements of $O^{omega}_n$ share the same $varphi$. Since $omega$ is arbitrary we have finished.
His surname is pure magic: Sivasankaranarayana.
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
add a comment |
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$begingroup$
Here it is an elementary argument inspired by Pillai$^{(*)}$: the point is that $nu_2(varphi(n))approxomega(n).$
Let $O^{omega}$ the set of odd natural numbers with at least $omega$ prime divisors and let $O^{omega}_n=O^{omega}cap[1,n]$.
$O^{omega}$ certainly has density $frac{1}{2}$ for any $omega$ - it is enough to invoke some elementary version of the PNT and basic sieving arguments. The totient function maps $O^{omega}_n$ into a subset of $[1,n]cap 2^{omega}mathbb{N}$, hence by the pigeonhole principle at least $(1-varepsilon_n)2^{omega-1}$ elements of $O^{omega}_n$ share the same $varphi$. Since $omega$ is arbitrary we have finished.
His surname is pure magic: Sivasankaranarayana.
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
add a comment |
$begingroup$
Here it is an elementary argument inspired by Pillai$^{(*)}$: the point is that $nu_2(varphi(n))approxomega(n).$
Let $O^{omega}$ the set of odd natural numbers with at least $omega$ prime divisors and let $O^{omega}_n=O^{omega}cap[1,n]$.
$O^{omega}$ certainly has density $frac{1}{2}$ for any $omega$ - it is enough to invoke some elementary version of the PNT and basic sieving arguments. The totient function maps $O^{omega}_n$ into a subset of $[1,n]cap 2^{omega}mathbb{N}$, hence by the pigeonhole principle at least $(1-varepsilon_n)2^{omega-1}$ elements of $O^{omega}_n$ share the same $varphi$. Since $omega$ is arbitrary we have finished.
His surname is pure magic: Sivasankaranarayana.
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
add a comment |
$begingroup$
Here it is an elementary argument inspired by Pillai$^{(*)}$: the point is that $nu_2(varphi(n))approxomega(n).$
Let $O^{omega}$ the set of odd natural numbers with at least $omega$ prime divisors and let $O^{omega}_n=O^{omega}cap[1,n]$.
$O^{omega}$ certainly has density $frac{1}{2}$ for any $omega$ - it is enough to invoke some elementary version of the PNT and basic sieving arguments. The totient function maps $O^{omega}_n$ into a subset of $[1,n]cap 2^{omega}mathbb{N}$, hence by the pigeonhole principle at least $(1-varepsilon_n)2^{omega-1}$ elements of $O^{omega}_n$ share the same $varphi$. Since $omega$ is arbitrary we have finished.
His surname is pure magic: Sivasankaranarayana.
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
Here it is an elementary argument inspired by Pillai$^{(*)}$: the point is that $nu_2(varphi(n))approxomega(n).$
Let $O^{omega}$ the set of odd natural numbers with at least $omega$ prime divisors and let $O^{omega}_n=O^{omega}cap[1,n]$.
$O^{omega}$ certainly has density $frac{1}{2}$ for any $omega$ - it is enough to invoke some elementary version of the PNT and basic sieving arguments. The totient function maps $O^{omega}_n$ into a subset of $[1,n]cap 2^{omega}mathbb{N}$, hence by the pigeonhole principle at least $(1-varepsilon_n)2^{omega-1}$ elements of $O^{omega}_n$ share the same $varphi$. Since $omega$ is arbitrary we have finished.
His surname is pure magic: Sivasankaranarayana.
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
answered Jan 4 at 0:09
Jack D'AurizioJack D'Aurizio
288k33280660
288k33280660
add a comment |
add a comment |
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$begingroup$
@Adam: what is the open problem you are talking about? I believe my analytic proof clearly shows/proves that $|varphi^{-1}(n)|$ is not bounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:51
1
$begingroup$
Carmicheal's conjecture states a different thing, i.e. that every value in the range of $varphi$ is attained at least twice. I am just stating that there are values attained with arbitrary multiplicity.
$endgroup$
– Jack D'Aurizio
Jan 3 at 22:56
1
$begingroup$
@JackD'Aurizio Ford's theorem states that for any $k$, there is $n$ with $|varphi^{-1}(n)|=k$, while your question only asks about $geq k$. I haven't looked up the proof of Ford's theorem, but I expect your problem can be solved in a much simpler way (and might have been known historically earlier)
$endgroup$
– Wojowu
Jan 3 at 23:17
1
$begingroup$
@Wojowu: fine. Here it is another overkill: by Pillai's theorem, the set of values taken by the $varphi$ function has density zero. It follows that $|varphi^{-1}(n)|$ is clearly unbounded.
$endgroup$
– Jack D'Aurizio
Jan 3 at 23:43
1
$begingroup$
@JackD'Aurizio typical that no one throws them a log too
$endgroup$
– Adam
Jan 4 at 0:16